OT: Interesting math problem...

[Fyron]

Ragnarok said:
The thing is the book he had didn't indicate if numbers could share the same value. It would be nice if they could as everyone said 00000+00000=00000 is a simple solution.

Unless this book is reknowned for giving trick questions, this is most likely not the solution.

[Ragnarok]

That is what I am thinking. As far as I know the book is just a regular math book for his grade. Unfortunately they cut the answers out of the back...

[Aiken]

Fyron, that's why I quickly deleted my post. But not quickly enough.

I remeber that I saw such tasks in a school. Common way to solve them is to rewrite example to:

10000*T+1000*R+100*I+10*E+D+10000*D+1000*R+100*I+1 0*V+E-10000*R-1000*I-100*V-10*E-T=0

With additional condition it should be quite easy to solve.

[spoon]

Got it:

17465 +
57496 =
-------
74961

T=1
R=7
I=4
E=6
D=5
V=9

I brute forced it, not having time for fancy shmancy formulae. (so, I could be wrong, very wrong!)

[Alneyan]

This problem is the proof I no longer know to do additions; while I figured out the V=9 part, I didn't think that I+I=4 could also mean I=7, and not only I=2.

[geoschmo]

Yay Spoon. I was getting ready to post that it was unsolvable. I had worked out that V was 9 and I was 4. But I got hung up thinking then that the only option for R was 2.

[Ragnarok]

Yay for spoon! Thanks a lot man! I really do appreciate it. Go have a few brewski's at the Cantina on my tab.

[vanbeke]

It looks like is correct
17465 +
57496 =
_____
74961

Brute force solving (sort of)
Assumptions are that all letters represent different numbers and standard notation is used (the first number in a string is not zero)
V must be either 0 (if D+E < 10) or 9 (if D+E > 9)
If V is 0 then I is 5; if V is 9 then I is 4
For either value of V, R must be either 2 or 7; but R can't be 2 since that would require both T and D to be 1
So R is 7
Then T+D = 6 I tried the 4 possibilities and the only one that worked is the form that I gave above.

[Will]

A more rigorous "proof", took me about five minutes before I went to class, posting it now after class:

Code:

---

 TRIED

+DRIVE

------

 RIVET



Assuming we disregard the trivial solution (all zeros), and 

force each letter to be distinct (ie for any x and y in 

{T,R,I,E,D,V}, the values x != y), and noticing that the 

carry can be at most 1 (9 + 9 + 1 = 19):



either E + V = E (V = 0)

    or E + V + 1 = E (V = 9)

disregard V = 0 as uninteresting, so V = 9

    and D + E > 10 and T < D, T < E

either I + I = V = 9

    or I + I + 1 = V = 9

I + I != 9, so I + I + 1 = V is true

so I = {4,9}, but V = 9, so I = 4

so R + R = I = 4 (no carry from previous digit)

then R = {2,7}

V = 9, I = 4, R = {2,7}

either T + D = R = 2

    or T + D + 1 = R = 7

assume there is no final carry

so if T + D = 2, then T = D = 1

then T + D + 1 = 7 = R

then T < (7 - D)

V = 9, I = 4, R = 7, T < (7 - D), T < E

try T = 1:

V = 9, I = 4, R = 7, T = 1, D = 5, E = 6

---

[Suicide Junkie]

Here is a tough one:
Fill in the long division using the blanks and the given numbers.
Code:

---

              _ _ 8 _

       _______________

_ _ _ | _ _ _ _ _ _ 5 

        _ _ _ _

        ========

          _ _ _ _

            _ _ _ 

          ========

            _ _ _ _

              _ _ _

            ========

              _ _ _ _

              _ _ _ _

              ========

                    0

---